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A142987
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a(1) = 1, a(2) = 10, a(n+2) = 10*a(n+1) + (n+1)*(n+2)*a(n).
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4
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1, 10, 106, 1180, 13920, 174600, 2330640, 33084000, 498646080, 7964020800, 134491276800, 2396163513600, 44942274316800, 885524502643200, 18293122632960000, 395457106963968000, 8930300425804800000
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| This is the case m = 5 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n+1)*(n+2)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.
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REFERENCES
| Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
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FORMULA
| a(n) = n!*p(n+1)*sum {k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (2*n^5+10*n^3+3*n)/15 = A069038(n). Recurrence: a(1) = 1, a(2) = 10, a(n+2) = 10*a(n+1)+(n+1)*(n+2)*a(n). The sequence b(n):= n!*p(n+1) satisfies the same recurrence with b(1) = 10, b(2) = 102. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(10 +1*2/(10 +2*3/(10 +3*4/(10 +...+(n-1)*n/10)))), for n >=2. The behaviour of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(10 +1*2/(10 +2*3/(10 +3*4/(10 +...+n*(n+1)/(10 +...))))) = 10*log(2) - 41/6, where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
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MAPLE
| p := n -> (2*n^5+10*n^3+3*n)/15: a := n -> n!*p(n+1)*sum ((-1)^(k+1)/(p(k)*p(k+1)), k = 1..n): seq(a(n), n = 1..20);
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CROSSREFS
| Cf. A069038, A142983, A142984, A142985, A142986.
Sequence in context: A087599 A068097 A190956 * A078192 A015589 A163192
Adjacent sequences: A142984 A142985 A142986 * A142988 A142989 A142990
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KEYWORD
| easy,nonn
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AUTHOR
| Peter Bala (pbala(AT)toucansurf.com), Jul 17 2008
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